Key Concepts and Formulas

• Equation of a Line in 3D: A line passing through a point $(x_0, y_0, z_0)$ with direction ratios $(a, b, c)$ can be represented in symmetric form as $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$. This can also be written in parametric form by setting it equal to a parameter, say $\lambda$, yielding $x = x_0 + a\lambda$, $y = y_0 + b\lambda$, $z = z_0 + c\lambda$.
• Distance of a Point from a Line in a Specific Direction: To find the distance of a point $P$ from a line $L_1$ along the direction of a vector $\vec{d}$, we construct a second line $L_2$ that passes through $P$ and is parallel to $\vec{d}$. The point of intersection $Q$ of $L_1$ and $L_2$ is the point on $L_1$ such that $PQ$ is parallel to $\vec{d}$. The required distance is then the length of the segment $PQ$.
• Distance Formula in 3D: The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$.

Step-by-Step Solution

Step 1: Identify Given Information and Understand the Problem Setup

We are asked to find the square of the distance of a given point $P$ from a line $L_1$ in the direction of a specific vector $\vec{d}$. This means we are not looking for the shortest (perpendicular) distance, but rather the length of a line segment $PQ$ where $Q$ is on $L_1$ and $PQ$ is parallel to $\vec{d}$.

• Given Point P: $P = \left( \frac{15}{7}, \frac{32}{7}, 7 \right)$.
• Given Line $L_1$ (Symmetric Form): $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$.
  • From this, we identify that $L_1$ passes through the point $(-1, -3, -5)$ and has a direction vector $\vec{v_1} = 3\hat{i} + 5\hat{j} + 7\hat{k}$.
• Given Direction Vector $\vec{d}$: $\vec{d} = \hat{i} + 4\hat{j} + 7\hat{k}$.
  • This is the direction along which we need to measure the distance.

The strategy is to find a point $Q$ on $L_1$ such that the line segment $PQ$ is parallel to $\vec{d}$. This implies $Q$ is the intersection of $L_1$ and a new line $L_2$ which passes through $P$ and is parallel to $\vec{d}$.

Step 2: Formulate the Parametric Equation of Line $L_1$

To find an intersection point, it's convenient to express the coordinates of any point on $L_1$ using a parameter. Let's set the symmetric form of $L_1$ equal to a parameter $\lambda$: $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} = \lambda$ From this, the coordinates of any point $Q(x_Q, y_Q, z_Q)$ on $L_1$ can be expressed as: $x_Q = 3\lambda - 1$ $y_Q = 5\lambda - 3$ $z_Q = 7\lambda - 5$ So, any point on $L_1$ is $Q(\lambda) = (3\lambda - 1, 5\lambda - 3, 7\lambda - 5)$.

Step 3: Formulate the Parametric Equation of Line $L_2$

Line $L_2$ passes through the given point $P = \left( \frac{15}{7}, \frac{32}{7}, 7 \right)$ and is parallel to the direction vector $\vec{d} = \hat{i} + 4\hat{j} + 7\hat{k}$. The symmetric form of $L_2$ is: $\frac{x - \frac{15}{7}}{1} = \frac{y - \frac{32}{7}}{4} = \frac{z - 7}{7}$ Let's set this equal to another parameter $\mu$: $\frac{x - \frac{15}{7}}{1} = \frac{y - \frac{32}{7}}{4} = \frac{z - 7}{7} = \mu$ From this, the coordinates of any point $R(x_R, y_R, z_R)$ on $L_2$ can be expressed as: $x_R = \mu + \frac{15}{7}$ $y_R = 4\mu + \frac{32}{7}$ $z_R = 7\mu + 7$ So, any point on $L_2$ is $R(\mu) = \left( \mu + \frac{15}{7}, 4\mu + \frac{32}{7}, 7\mu + 7 \right)$.

Step 4: Find the Intersection Point $Q$

The point $Q$ we are looking for is the intersection of $L_1$ and $L_2$. At this point, the coordinates from both parametric forms must be equal. We equate the corresponding coordinates to find the values of $\lambda$ and $\mu$.

1. Equating x-coordinates: $3\lambda - 1 = \mu + \frac{15}{7}$ $3\lambda - \mu = 1 + \frac{15}{7} \implies 3\lambda - \mu = \frac{7 + 15}{7} \implies 3\lambda - \mu = \frac{22}{7}$ (Equation A)

2. Equating y-coordinates: $5\lambda - 3 = 4\mu + \frac{32}{7}$ $5\lambda - 4\mu = 3 + \frac{32}{7} \implies 5\lambda - 4\mu = \frac{21 + 32}{7} \implies 5\lambda - 4\mu = \frac{53}{7}$ (Equation B)

3. Equating z-coordinates: $7\lambda - 5 = 7\mu + 7$ $7\lambda - 7\mu = 7 + 5 \implies 7\lambda - 7\mu = 12 \implies \lambda - \mu = \frac{12}{7}$ (Equation C)

Now we solve the system of linear equations for $\lambda$ and $\mu$. We can use Equations A and C, as they are simpler. From Equation C, we can express $\mu$ in terms of $\lambda$: $\mu = \lambda - \frac{12}{7}$

Substitute this expression for $\mu$ into Equation A: $3\lambda - \left( \lambda - \frac{12}{7} \right) = \frac{22}{7}$ $3\lambda - \lambda + \frac{12}{7} = \frac{22}{7}$ $2\lambda = \frac{22}{7} - \frac{12}{7}$ $2\lambda = \frac{10}{7}$ $\lambda = \frac{5}{7}$

Now, substitute the value of $\lambda$ back into the expression for $\mu$: $\mu = \frac{5}{7} - \frac{12}{7}$ $\mu = -\frac{7}{7}$ $\mu = -1$

(Optional Verification): We can check these values in Equation B: $5\left(\frac{5}{7}\right) - 4(-1) = \frac{25}{7} + 4 = \frac{25 + 28}{7} = \frac{53}{7}$. This matches the right-hand side of Equation B, so our values for $\lambda$ and $\mu$ are correct.

Finally, we find the coordinates of the intersection point $Q$ by substituting $\lambda = \frac{5}{7}$ into the parametric form of $L_1$: $x_Q = 3\left(\frac{5}{7}\right) - 1 = \frac{15}{7} - \frac{7}{7} = \frac{8}{7}$ $y_Q = 5\left(\frac{5}{7}\right) - 3 = \frac{25}{7} - \frac{21}{7} = \frac{4}{7}$ $z_Q = 7\left(\frac{5}{7}\right) - 5 = 5 - 5 = 0$ So, the intersection point is $Q = \left( \frac{8}{7}, \frac{4}{7}, 0 \right)$.

Step 5: Calculate the Distance $PQ$

Now we calculate the distance between point $P = \left( \frac{15}{7}, \frac{32}{7}, 7 \right)$ and point $Q = \left( \frac{8}{7}, \frac{4}{7}, 0 \right)$ using the 3D distance formula: $PQ = \sqrt{\left( x_P - x_Q \right)^2 + \left( y_P - y_Q \right)^2 + \left( z_P - z_Q \right)^2}$ $PQ = \sqrt{\left( \frac{15}{7} - \frac{8}{7} \right)^2 + \left( \frac{32}{7} - \frac{4}{7} \right)^2 + \left( 7 - 0 \right)^2}$ $PQ = \sqrt{\left( \frac{7}{7} \right)^2 + \left( \frac{28}{7} \right)^2 + \left( 7 \right)^2}$ $PQ = \sqrt{(1)^2 + (4)^2 + (7)^2}$ $PQ = \sqrt{1 + 16 + 49}$ $PQ = \sqrt{66}$

Step 6: Square the Distance

The question asks for the square of the distance. $(PQ)^2 = \left( \sqrt{66} \right)^2 = 66$

Common Mistakes & Tips

• Perpendicular vs. Directional Distance: The most common mistake is to confuse this problem with finding the shortest (perpendicular) distance from a point to a line. Always pay close attention to the phrase "in the direction of the vector".
• Fractional Arithmetic: The coordinates involve fractions. Be meticulous with arithmetic involving fractions to avoid calculation errors, especially when solving the system of equations.
• Systematic Approach: Break down the problem into smaller, manageable steps (defining lines, finding intersection, calculating distance). This reduces complexity and helps in identifying errors.

Summary

To find the distance of a point from a line in a given direction, we constructed a second line through the point, parallel to the specified direction. We then found the intersection point of these two lines. The distance between the given point and this intersection point is the required distance. After calculating this distance as $\sqrt{66}$, we squared it as requested by the problem.

The final answer is $\boxed{\text{66}}$, which corresponds to option (A).